# Quick questions and answers

#

**1**lives in a giant shoe

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**Quick questions and answers**

I know I'm not a mod here, but this might alleviate some of the thread strain around these parts. In this thread you can ask quick questions that don't really warrant a thread. Something like "what do you guys think of *****". That would be a good question to ask in here. Also "do you guys use pedals on your bikes?" would also be appropriate. If this thread relieves the front page from two threads a day then it will be well worth keeping around.

#

**3**King Among Runaways

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I really don't get why so many people are so anal about "unnecessary" or "redundant" threads.

Why can't people just ignore the threads that don't interest them and allow those who want to start their own thread to do so - regardless of the topic.

I don't understand the need to "relieve the front page from two threads a day."

So, what do you all think of that?

Why can't people just ignore the threads that don't interest them and allow those who want to start their own thread to do so - regardless of the topic.

I don't understand the need to "relieve the front page from two threads a day."

So, what do you all think of that?

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**4**F*** Corporate Beer

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Some people need to type less and ride more.

I tried to ride today, but the 4+ inches of snow put a damper on that.

I need some skis now...or a Puglsey!

I tried to ride today, but the 4+ inches of snow put a damper on that.

I need some skis now...or a Puglsey!

#

**7**WTF?

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yeah, just ignore the threads you don't like...

i'm not sure how to put this philosophically...people waste their time writing replies because they CAN... just like i'm doing now.

i'm not sure how to put this philosophically...people waste their time writing replies because they CAN... just like i'm doing now.

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**8**If you don't look good...

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Threads that you have no knowledge to contribute to or have anothing to do with... DO NOT REPLY.

Easy!

ps. yes... kinda like what I just did... don't do that!

Easy!

ps. yes... kinda like what I just did... don't do that!

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**9**ass hatchet

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Check out my new "I Already Knew That" thread. It seeks to alieviate some of the stress of the "Quick Question" thread by inviting people to shed light on the fact that they already know the answer to whatever question you intend to ask (as oppoesed to actually answering 'em). Perhaps the mods will make it a "sticky."

#

**10**hang up your boots

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I've always had some quick questions in mind, now i have chance to ask them:

• If a stripper gets breast implants can she write it off on her taxes as a business expense?

• How do "do not walk on grass" signs get there?

• If there was a crumb on the table and you cut it in half, would you have two crumbs or two halves of a crumb?

• Do illiterate people get the full effect of Alphabet soup?

• In that song, she'll be coming around the mountain, who is she?

and lastly

• Why do all superheroes wear spandex?

• If a stripper gets breast implants can she write it off on her taxes as a business expense?

• How do "do not walk on grass" signs get there?

• If there was a crumb on the table and you cut it in half, would you have two crumbs or two halves of a crumb?

• Do illiterate people get the full effect of Alphabet soup?

• In that song, she'll be coming around the mountain, who is she?

and lastly

• Why do all superheroes wear spandex?

#

**11**Iguana Subsystem

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Originally Posted by

**ostro**I've always had some quick questions in mind, now i have chance to ask them:

• If a stripper gets breast implants can she write it off on her taxes as a business expense?

• How do "do not walk on grass" signs get there?

• If there was a crumb on the table and you cut it in half, would you have two crumbs or two halves of a crumb?

• Do illiterate people get the full effect of Alphabet soup?

• In that song, she'll be coming around the mountain, who is she?

and lastly

• Why do all superheroes wear spandex?

• If a stripper gets breast implants can she write it off on her taxes as a business expense?

• How do "do not walk on grass" signs get there?

• If there was a crumb on the table and you cut it in half, would you have two crumbs or two halves of a crumb?

• Do illiterate people get the full effect of Alphabet soup?

• In that song, she'll be coming around the mountain, who is she?

and lastly

• Why do all superheroes wear spandex?

- yes
- the rule doesn't go into effect until the feet of the installer are off the grass
- two crumbs
- no
- sally the mule (don't ask)
- because it's easy to get stains out of spandex

next.

#

**12**hang up your boots

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ok...

• If you die and you have a broken leg do they take the cast off?

• Since there is a rule that states "i" before "e" except after "c", isnt "science" spelled wrong?

• Why isn't the word 'gullible' in the dictionary?

•*Why did Superman wear his briefs on the outside of his tights?

• How many licks does it take to get to the center of a tootsie pop?

• Why do British people never sound British when they sing?

• If you die and you have a broken leg do they take the cast off?

• Since there is a rule that states "i" before "e" except after "c", isnt "science" spelled wrong?

• Why isn't the word 'gullible' in the dictionary?

•*Why did Superman wear his briefs on the outside of his tights?

• How many licks does it take to get to the center of a tootsie pop?

• Why do British people never sound British when they sing?

#

**13**Iguana Subsystem

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Originally Posted by

**ostro**ok...

• If you die and you have a broken leg do they take the cast off?

• Since there is a rule that states "i" before "e" except after "c", isnt "science" spelled wrong?

• Why isn't the word 'gullible' in the dictionary?

•*Why did Superman wear his briefs on the outside of his tights?

• How many licks does it take to get to the center of a tootsie pop?

• Why do British people never sound British when they sing?

• If you die and you have a broken leg do they take the cast off?

• Since there is a rule that states "i" before "e" except after "c", isnt "science" spelled wrong?

• Why isn't the word 'gullible' in the dictionary?

•*Why did Superman wear his briefs on the outside of his tights?

• How many licks does it take to get to the center of a tootsie pop?

• Why do British people never sound British when they sing?

- yes
- it's spelled correctly, but english doesn't follow rules very well
- nice try
- 'cause he's an alien
- no one knows, because you always crunch them up before you get there
- they do sound british, it just so happens that recieved pronunciation sounds north american when sung.

#

**17**hang up your boots

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damn, the quick question thread works.

What is the geometrical meaning of the central extension of the algebra of diffeomorphisms of a circle?

What is the geometrical meaning of the central extension of the algebra of diffeomorphisms of a circle?

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**19**hang up your boots

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Originally Posted by

**dolface**of the whole group, or for a given circle?

#

**20**Iguana Subsystem

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ok, i haven't looked at the problem in a long time, so all i've got is a general explanation (you can plug in the numbers for whatever circle you have).

(if you don't care about the math [which sorta makes it pointless] this is a very basic description: there is a lie algebra called the virasoro algebra; its almost just the lie algebra of the group of diffeomorphisms of the circle, but it's actually just one dimension bigger, being a "central extension" thereof; projective representations of the lie algebra of the group of diffeomorphisms of the circle correspond to honest representations of the virasoro algebra.)

i'm not gonna re-type my notes, but this post (from google groups) covers it pretty well:

"In 2D the conformal group is infinite-dimensional - it

consists of transformations which are analytic in z = x+iy and z* =

x-iy. If we concentrate on the former, we see that an infinitesimal

conformal transformation is generated by

L_m = z^{m+1} d/dz,

and they obey the algebra

[L_m, L_n] = (n-m) L_{m+n}.

However, it is easy to see that this is the algebra of vector fields

(= infinitesimal diffeomorphisms) on the circle, vect(1). If the

circle coordinate is x, the generators are

L_m = -i exp(imx) d/dx.

vect(1) has a central extension, known as the Virasoro algebra:

[L_m, L_n] = (n-m) L_{m+n} - c/12 (m^3 - m) delta_{m+n,0},

where c is a c-number known as the central charge or conformal

anomaly. This means that the Virasoro algebra is still a Lie

algebra - anti-symmetry and the Jacobi identities still hold. The

term linear in m is unimportant, because it can be removed by a

redefinition of L_0.

The m^3 term is a genuine quantum effect, which simply is there when

you quantize a string. When string theorists criticize Thiemann's

LQG string, they are basically complaining that he does not get

this term, which simply must be there.

There is some confusion about anomaly freedom here, because at the

end people want to eliminate the conformal symmetry. The nice way to

do this is to introduce a ghost pair b_m, c_n, satisfying fermionic

brackets

{ b_m, c_n } = delta_{m+n,0}.

One can now write down the BRST operator, which looks something

like (double dots denote normal ordering)

Q = sum_m :L_{-m} c_m: + 1/2 sum_{m,n} (m-n) :b_{-m-n} c_m c_n:

If the BRST operator is nilpotent, Q^2 = 0, we can identify

physical states with BRST cohomology. A state is physical if it

is BRST closed, Q|phys> = 0, and two states are equivalent if

they differ by a BRST exact term,

|phys> ~ |phys'> if Q( |phys> - |phys'> ) = 0.

It turns out that the ghost has central charge c = -26, so the BRST

operator is nilpotent if the L_m's have c = 26; this is where the 26

dimensions of they bosonic string comes from. However, the important

thing from my viewpoint is not that the end result is anomaly free,

but that an anomaly exists, even if only in intermediate

calculations. Thiemann does not have an anomaly even intermediately,

and from this string theorists (and myself) conclude that LQG is

wrong.

Let us now return to the math. A lowest-weight representation (LWR)

of the Virasoro algebra is characterized by a lowest-weight state

|h,c> satisfying

L_0 |h,c> = h |h,c>,

L_{-m} |h,c> = 0, for all -m < 0.

It is known that the only unitary LWR of vect(1) is the trivial one.

However, the Virasoro algebra has many unitary LWRs: the discrete

series with 0 <= c < 1, where

c = 1 - 6/m(m+1), m positive integer and

h = h_{p,q}(c) = (pm^2 - q(m+1)^2) / 4m(m+1)

(or something similar, I'm quoting from memory), and also all c > 1,

h > 0. Anyway, the important thing is that the only acceptable value

of (h,c) with c = 0 is h = 0 - this is the trivial representation.

That Thiemann obtains a non-trivial unitary representation of the

1D diffeomorphism group with c = 0 is thus very strange. It is

hard to see how that could be compatible with quantum theory.

The Virasoro algebra can be generalized to several dimensions -

an extension of the diffeomorphism algebra on the N-dimensional

torus, say. The generators are

L_k(m) = i exp(im.x) d/dx^k,

where m = (m_i) and m.x = m_i x^i and I use the summation

convention. The algebra depends on two parameters (abelian charges)

c_1 and c_2,

[L_i(m), L_j(n)] = n_i L_j(m+n) - m_j L_i(m+n)

+ (c_1 m_j n_i + c_2 m_i n_j) m_k S^k(m+n),

[L_i(m), S^j(n)] = n_i S^j(m+n) + delta^j_i m_k S^k(m+n),

[S^i(m), S^j(n)] = 0,

m_k S^k(m) = 0.

This is the Virasoro algebra in 1D because the last condition then

becomes m S(m) = 0, which only has the solution S(m) ~ delta(m).

The Virasoro extension is not central (does not commute with

everything) except in 1D. It is straightforward but somewhat tedious

to check that these relations indeed do define a Lie algebra.

Just as the Virasoro algebra is anomalous when c != 0, its higher-

dimensional sibling is anomalous unless both c_1 = c_2 = 0. And just

as for the Virasoro algebra, there is no non-trivial, unitary LWRs

unless the algebra is anomalous. The correct definition of

lowest-weight is more subtle in several dimensions. Let me just say

that the right definition does not introduce any anisotropy."

(if you don't care about the math [which sorta makes it pointless] this is a very basic description: there is a lie algebra called the virasoro algebra; its almost just the lie algebra of the group of diffeomorphisms of the circle, but it's actually just one dimension bigger, being a "central extension" thereof; projective representations of the lie algebra of the group of diffeomorphisms of the circle correspond to honest representations of the virasoro algebra.)

i'm not gonna re-type my notes, but this post (from google groups) covers it pretty well:

"In 2D the conformal group is infinite-dimensional - it

consists of transformations which are analytic in z = x+iy and z* =

x-iy. If we concentrate on the former, we see that an infinitesimal

conformal transformation is generated by

L_m = z^{m+1} d/dz,

and they obey the algebra

[L_m, L_n] = (n-m) L_{m+n}.

However, it is easy to see that this is the algebra of vector fields

(= infinitesimal diffeomorphisms) on the circle, vect(1). If the

circle coordinate is x, the generators are

L_m = -i exp(imx) d/dx.

vect(1) has a central extension, known as the Virasoro algebra:

[L_m, L_n] = (n-m) L_{m+n} - c/12 (m^3 - m) delta_{m+n,0},

where c is a c-number known as the central charge or conformal

anomaly. This means that the Virasoro algebra is still a Lie

algebra - anti-symmetry and the Jacobi identities still hold. The

term linear in m is unimportant, because it can be removed by a

redefinition of L_0.

The m^3 term is a genuine quantum effect, which simply is there when

you quantize a string. When string theorists criticize Thiemann's

LQG string, they are basically complaining that he does not get

this term, which simply must be there.

There is some confusion about anomaly freedom here, because at the

end people want to eliminate the conformal symmetry. The nice way to

do this is to introduce a ghost pair b_m, c_n, satisfying fermionic

brackets

{ b_m, c_n } = delta_{m+n,0}.

One can now write down the BRST operator, which looks something

like (double dots denote normal ordering)

Q = sum_m :L_{-m} c_m: + 1/2 sum_{m,n} (m-n) :b_{-m-n} c_m c_n:

If the BRST operator is nilpotent, Q^2 = 0, we can identify

physical states with BRST cohomology. A state is physical if it

is BRST closed, Q|phys> = 0, and two states are equivalent if

they differ by a BRST exact term,

|phys> ~ |phys'> if Q( |phys> - |phys'> ) = 0.

It turns out that the ghost has central charge c = -26, so the BRST

operator is nilpotent if the L_m's have c = 26; this is where the 26

dimensions of they bosonic string comes from. However, the important

thing from my viewpoint is not that the end result is anomaly free,

but that an anomaly exists, even if only in intermediate

calculations. Thiemann does not have an anomaly even intermediately,

and from this string theorists (and myself) conclude that LQG is

wrong.

Let us now return to the math. A lowest-weight representation (LWR)

of the Virasoro algebra is characterized by a lowest-weight state

|h,c> satisfying

L_0 |h,c> = h |h,c>,

L_{-m} |h,c> = 0, for all -m < 0.

It is known that the only unitary LWR of vect(1) is the trivial one.

However, the Virasoro algebra has many unitary LWRs: the discrete

series with 0 <= c < 1, where

c = 1 - 6/m(m+1), m positive integer and

h = h_{p,q}(c) = (pm^2 - q(m+1)^2) / 4m(m+1)

(or something similar, I'm quoting from memory), and also all c > 1,

h > 0. Anyway, the important thing is that the only acceptable value

of (h,c) with c = 0 is h = 0 - this is the trivial representation.

That Thiemann obtains a non-trivial unitary representation of the

1D diffeomorphism group with c = 0 is thus very strange. It is

hard to see how that could be compatible with quantum theory.

The Virasoro algebra can be generalized to several dimensions -

an extension of the diffeomorphism algebra on the N-dimensional

torus, say. The generators are

L_k(m) = i exp(im.x) d/dx^k,

where m = (m_i) and m.x = m_i x^i and I use the summation

convention. The algebra depends on two parameters (abelian charges)

c_1 and c_2,

[L_i(m), L_j(n)] = n_i L_j(m+n) - m_j L_i(m+n)

+ (c_1 m_j n_i + c_2 m_i n_j) m_k S^k(m+n),

[L_i(m), S^j(n)] = n_i S^j(m+n) + delta^j_i m_k S^k(m+n),

[S^i(m), S^j(n)] = 0,

m_k S^k(m) = 0.

This is the Virasoro algebra in 1D because the last condition then

becomes m S(m) = 0, which only has the solution S(m) ~ delta(m).

The Virasoro extension is not central (does not commute with

everything) except in 1D. It is straightforward but somewhat tedious

to check that these relations indeed do define a Lie algebra.

Just as the Virasoro algebra is anomalous when c != 0, its higher-

dimensional sibling is anomalous unless both c_1 = c_2 = 0. And just

as for the Virasoro algebra, there is no non-trivial, unitary LWRs

unless the algebra is anomalous. The correct definition of

lowest-weight is more subtle in several dimensions. Let me just say

that the right definition does not introduce any anisotropy."

#

**23**bannned

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dol, that made me nauseous.

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**24**Iguana Subsystem

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Originally Posted by

**steaktaco**dol, that made me nauseous.

i need another beer

#

**25**i am sure that i hate you

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dolface, if you didnt copy/paste that, you need to get out on your bike

__________________

putting the pi back in

putting the pi back in

**pirate**!
It’s an upstanding member of the solar system

Apply the laws of earth and make it a victim

Of Proposition 187

Apply the laws of earth and make it a victim

Of Proposition 187